(B) The distance between two successive nodes in a stationary wave is $\frac{\lambda}{2}$.For $6$ successive nodes,there are $5$ such intervals.Theref

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(B) The distance between two successive nodes in a stationary wave is $\frac{\lambda}{2}$.For $6$ successive nodes,there are $5$ such intervals.Therefore,the total distance is $d = 5 \times \frac{\lambda}{2} = 85 \text{ cm}$.Solving for wavelength $\lambda$: $\lambda = \frac{2 \times 85}{5} = 34 \text{ cm} = 0.34 \text{ m}$.The speed of sound $v$ is given by $v = f \lambda$,where $f = 1000 \text{ Hz}$.$v = 1000 \times 0.34 = 340 \text{ m/s}$.

Class 11 Physics Waves and Sound Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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On producing waves of frequency $1000 \text{ Hz}$ in a Kundt's tube,the total distance between $6$ successive nodes is $85 \text{ cm}$. The speed of sound in the gas filled in the tube is ..... $m/s$.

A
$330$
B
$340$
C
$350$
D
$300$

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Waves and Sound

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Longitudinal Stationary Waves (Organ Pipes) and Resonance Tube

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DifficultMHT CET 2024

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Tube $A$ has both ends open while tube $B$ has one end closed,otherwise they are identical. The ratio of the fundamental frequency of tube $A$ to tube $B$ is

MediumAIEEE 2002

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On producing waves of frequency $1000 \text{ Hz}$ in a Kundt's tube,the total distance between $6$ successive nodes is $85 \text{ cm}$. The speed of sound in the gas filled in the tube is ..... $m/s$.

Similar Questions

$A$ pipe open at both ends has a fundamental frequency '$f$' in air. The pipe is dipped in water, so that $\frac{2}{3}$ of its length is in water. Now, the fundamental frequency of the air column is:

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$A$ tuning fork of frequency $340 \ Hz$ is vibrated just above a tube of $120 \ cm$ height. Water is slowly poured in the tube. What is the minimum height of water necessary for resonance (in $cm$)?

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